Why is the velocity of a jump on Mars the same as on Earth?

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  • #1
HAF
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Homework Statement


Good evening,
I have a homework where the question is "How high will person jump on Mars?"


Homework Equations


v2 = 2.amars.h1
V2 = 2.g.h2

The Attempt at a Solution


I know everything to solve it. I can solve it by h1 / h2 . In this created equation the only unknown is h1 so it's all cool. But why can I assume that v1 and V1 are the same? Why jump velocity on Mars is the same as on Earth? Aren't there any changes? For example lower strength of muscles due to different weight(force)? I know that maybe my question is completely stupid and I'm very sorry about it. I'm happy that I've atleast tried to find it out but I think that I can't figure out why the velocities are the same without your help.
 
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Answers and Replies

  • #2
jbriggs444
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But why can I assume that v1 and V1 are the same?
Because it is homework and one is expected to use models that are simplified to the point of being incorrect.
 
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  • #3
HAF
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Because it is homework and one is expected to use models that are simplified to the point of being incorrect.
Thank you very much sir. Now everything makes sense to me. I really appreciate it!
 
  • #4
haruspex
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Why jump velocity on Mars is the same as on Earth?
The assumption is that the work the legs can do is the same. There is also the issue of how the height of jump is defined, but vertical displacement of mass centre (from the crouch position) is the biophysical standard. So it is not really about take off speed. How is that defined? At the moment the feet leave the ground, the Martian jumper would have a greater speed, having done less work against gravity already.

Of course, it is not really true that the same work will be done. Muscles have a couple of limits - max force and max contraction rate - with a nonlinear response along the way.
 
  • #5
HAF
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The assumption is that the work the legs can do is the same. There is also the issue of how the height of jump is defined, but vertical displacement of mass centre (from the crouch position) is the biophysical standard. So it is not really about take off speed. How is that defined? At the moment the feet leave the ground, the Martian jumper would have a greater speed, having done less work against gravity already.

Of course, it is not really true that the same work will be done. Muscles have a couple of limits - max force and max contraction rate - with a nonlinear response along the way.
Aha. So for correctness should I solve it from equation m.g.h(1) = m.a.h(2) right?
 
  • #6
PeroK
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There is also the issue of how the height of jump is defined, but vertical displacement of mass centre (from the crouch position) is the biophysical standard.
Or, you could refer to the rules of the Olympic Standing Jump, from 1912:

 
  • #7
haruspex
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Aha. So for correctness should I solve it from equation m.g.h(1) = m.a.h(2) right?
That's how I would approach it.
 

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